The magnetic moment of an octahedral homolept | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. For an octahedral complex,the magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,whe

Vedclass · Accepted Answer

$NCS^{-}$

Vedclass · Answer

(C) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. For an octahedral complex,the magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. $A$ magnetic moment of $5.9 \ BM$ corresponds to $n = 5$ unpaired electrons. This indicates that the complex is a high-spin complex,which occurs when the ligand is a weak field ligand. Among the given options,$NCS^{-}$ is a weak field ligand,while $ethylenediamine$,$CN^{-}$,and $CO$ are strong field ligands that would cause pairing of electrons. Therefore,the complex is $[Mn(NCS)_6]^{4-}$.

The magnetic moment of an octahedral homoleptic $Mn(II)$ complex is $5.9 \ BM$. The suitable ligand for this complex is:

Similar Questions

Which coordination number is associated with geometric isomerism in coordination compounds?

The theoretical value of the magnetic moment of the complex formed by the reaction of $Fe^{3+}$ ion with $CN^{-}$ ligand is .......... $B.M.$

What type of isomerism is exhibited by $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$?

Which one of the following complexes will exhibit the least paramagnetic behaviour?
[Atomic number: $Cr=24, Mn=25, Fe=26, Co=27$]

Complexes $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ can be distinguished by

Vedclass Test Series

Exam Paper Generator

Online Exam Module